Problem: What is the slope of the line tangent to $f(x) = x^{2}+3x-3$ at $x = -2$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{((x+h)^{2}+3(x+h)-3) - (x^{2}+3x-3)}{h}$ $ = \lim_{h \to 0} \frac{(x^{2}+2x h+h^{2}+3(x+h)-3) - (x^{2}+3x-3)}{h}$ $ = \lim_{h \to 0} \frac{x^{2}+2(x h)+h^{2}+3x+3h-3-x^{2}-3x+3}{h}$ $ = \lim_{h \to 0} \frac{2(x h)+h^{2}+3h}{h}$ $ = \lim_{h \to 0} 2x+h+3$ $ = 2x+3$ $ = (2)(-2)+3$ $ = -1$